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XO
29 tháng 1 2020
Ta có :\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)=2\times\left(\frac{1}{2}-\frac{1}{16}\right)=\frac{7}{8}\)
LQ
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 12 2023
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) +.. . + \(\dfrac{1}{120}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{120}\))
A = 2.( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + ... + \(\dfrac{1}{240}\))
A = 2.( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{15.16}\))
A =2 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{15}\) - \(\dfrac{1}{16}\))
A = 2.( \(\dfrac{1}{2}\) - \(\dfrac{1}{16}\))
A = 2.\(\dfrac{7}{16}\)
A = \(\dfrac{7}{8}\)
5 tháng 7 2018
Chỉ làm bài khó thôi nhé:::::::::::::::
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2018}\Rightarrow x+1=2018\Rightarrow x=2017\)
6 tháng 7 2016
b1/A=25/1.6+25/6.11+25/11.16+....+25/41.46
=5.(5/1.6+5/6.11+5/11.16+...+5/41.46)
=5.(1/1-1/6+1/6-1/11+1/11-1/16+....+1/41-1/46)
=5.(1/1-1/46)
=5.45/46
=225/46
HT
3 tháng 7 2018
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
LT
7 tháng 7 2018
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{120}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{240}\right)\)
\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{15\cdot16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{16}\right)\)
\(=2\left(\frac{8}{16}-\frac{1}{16}\right)\)
\(=2\cdot\frac{7}{16}\)
\(=\frac{7}{8}\)
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{120}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{240}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{240}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{16}\right)\)
\(=\frac{7}{8}\)